Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main {

    static double[][] dp;

    static double calculate(double[] arr, int count, int pos){
        if(count == 0)
            return 1;
        if(pos >= arr.length)
            return 0;
        if(dp[count][pos] != -1)
            return dp[count][pos];

        dp[count][pos] = (1-arr[pos])*calculate(arr,count,pos+1) + (arr[pos]*calculate(arr,count-1,pos+1));

        return dp[count][pos];
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int heads = (n/2);
        double[] arr = new double[n];
        dp = new double[n+1][n+1];

        for(int i  =0; i < n+1; i ++){
            for (int j =0; j < n+1; j++){
                dp[i][j] = -1;
            }
        }

        for(int i=0; i < n; i++){
            arr[i] = sc.nextDouble();
        }

        double ans = 0;

//        for(int i =  heads+1 ; i <= n; i++){
//            ans += calculate(arr,i,0);
//        }

        System.out.println(calculate(arr,heads+1,0));
    }
}